Question:
If the fourth term in the binomial expansion of
$\left(\frac{2}{x}+x^{\log _{8} x}\right)^{6}(x>0)$ is $20 \times 8^{7}$, then a value of
$x$ is :
Correct Option: , 2
Solution:
$\mathrm{T}_{4}=\mathrm{T}_{3+1}=\left(\begin{array}{l}6 \\ 3\end{array}\right)\left(\frac{2}{\mathrm{x}}\right)^{3} \cdot\left(\mathrm{x}^{\log _{\mathrm{s}} \mathrm{x}}\right)^{3}$
$20 \times 8^{7}=\frac{160}{x^{3}} \cdot x^{3 \log _{8} x}$
$8^{6}=x^{\log _{2} x}-3$
$2^{18}=x^{\log _{2} x-3}$
$\Rightarrow 18=\left(\log _{2} x-3\right)\left(\log _{2} x\right)$
Let $\log _{2} x=t$
$\Rightarrow t^{2}-3 t-18=0$
$\Rightarrow(t-6)(t+3)=0$
$\Rightarrow t=6,-3$
$\log _{2} x=6 \Rightarrow x=2^{6}=8^{2}$
$\log _{2} x=-3 \Rightarrow x=2^{-3}=8^{-1}$